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![](https://rs.olm.vn/images/avt/0.png?1311)
A=1.5.(3.2)+2.10.(6.2)+3.15.(9.2)+4.20.(12.2)+5.25.(15.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=1.5.3+2.10.6+3.15.9+4.20.12+5.25.15(2.2.2.2.2)
1.3.5+2.6.10+3.9.15+4.12.20+5.15.25
A=2.2.2.2.2
A=32
\(\frac{1\cdot3\cdot5\cdot2+2\cdot10\cdot6\cdot2+3\cdot15\cdot9\cdot2+4\cdot20\cdot12\cdot2+5\cdot25\cdot15\cdot2}{1\cdot3\cdot5+2\cdot10\cdot6+3\cdot15\cdot9+4\cdot20\cdot12+5\cdot25\cdot15 }\)
\(2\cdot2\cdot2\cdot2\cdot2=2^5\)
\(=32\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=-\frac{1}{3}\)
Ta có:\
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(A=-\frac{2}{6}=-\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(M=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\)
\(M=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(M=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(M=\frac{-2}{6}=\frac{-1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{7.2^{32}.3^8.5^4-2^9.3^9.2^8.5^4}{2^{10}.3^{10}.2^5.5^5-7.3^9.4^8.5^4}=\frac{7.2^{32}.3^8.5^4-2^{17}.3^9.5^4}{2^{15}.3^{10}.5^5-7.3^9.2^{16}.5^4}\)
\(\frac{2^{17}.3^8.5^4\left(2^5.7-1\right)}{2^{15}.3^9.5^4\left(3.5-7.2\right)}=\frac{2^{17}.3^8.5^4\left(32.7-1\right)}{2^{15}.3^9.5^4\left(15-14\right)}\)